Answer the following :
The pH of rainwater collected in a certain region of Maharashtra on a particular day was 5.1. Calculate the H+ ion concentration of the rainwater and its percent dissociation.
Solution
Given: pH of rainwater = 5.1
To find:
i. H+ ion concentration
ii. Percent dissociation
Formula:
i. pH = -log10[H3O+]
ii. Percent dissociation = α × 100
Calculation: From the formula (i),
pH = -log10[H3O+]
∴ log10[H3O+] = -5.1
= -5 - 0.1 + 1 - 1
= (-5 - 1) + 1 - 0.1
= -6 + 0.9 = `bar(6).9`
∴ [H3O+] = Antilog10[`bar(6).9`]
= 7.943 × 10-6 M
Considering that the pH of rainwater is due to the dissociation of a monobasic strong acid (HA), we have
\[\ce{HA_{(aq)} + H2O_{(l)} -> H3O^+_{ (aq)} + A^-_{ (aq)}}\]
∴ [H3O+] = α
α = `7.943 xx 10^-6`
From formula (ii),
Percent dissociation = `7.943 xx 10^-6 xx 100 = 7.943 xx 10^-4`
i. H+ ion concentration is `7.943 xx 10^-6` M
ii. Percent dissociation is `7.943 xx 10^-4`.
